I don't know if this question is appropriate to this site, but I posted here without an answer, so I tried this alternative.

Given a $2$-form $\omega$ on a manifold $M$, let us denote by $N$ the kernel of $\omega$, i.e. $N:=\{u\in TM : \omega(u,\cdot)=0\}$. Their Proposition 5.1.2 shows that if $\omega$ has constant rank (and is closed) then $N$ is a tangent distribution on $M$ (and completely integrable).

In the following remark they say that ``the reader can easily prove the converse of the previous conclusion''. While I understand that $N$ is a tangent distribution if and only if $\omega$ has constant rank. Instead I think that, for $\omega$ of constant rank, $N$ can be completely integrable even if $\omega$ is not closed, (e.g. $\omega=e^z dx\wedge dy$).

Starting from this consideration I have asked myself a question:

Given a $\Omega\in\mathcal{A}^p(M)$, with $p>1$, whose rank is constant, let us define its kernel $N$ as above. Evidently $N$ is a tangent distribution on $M$, and I find it is completely integrable at least when there exists a $1$-form $\phi$ such that $d\Omega=\phi\wedge\Omega$. Clearly, if $\Omega$ is decomposable then the last condition is even necessary.

My question (**edited after the comment of Willie Wong**):

Is this last condition (the ``divisibility'' of $d\Omega$ by $\Omega$) necessary for the complete integrability of $N$ even when $\Omega$ is not decomposable? (Using Frobenius' Theorem I understand the case $p=1$, but what about the case $p>1$?.)

Any suggestion and\or counterexample are welcome.